Engineering Mathematics: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Limits

Direct substitution
try x = a first; 0/0 means simplify
Factor and cancel
xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + a·xⁿ⁻² + … + aⁿ⁻¹)
Conjugate
(√x − √a)(√x + √a) = x − a
Limits at infinity
divide by the highest power of x; equal degrees give the ratio of leading coefficients
Special limit
limx→∞ (1 + a/x)bx = eab
L'Hôpital's rule
lim f/g = lim f′/g′ (0/0 or ∞/∞)

Derivatives

Power rule
(xⁿ)′ = n·xⁿ⁻¹
Product rule
(uv)′ = u′v + uv′
Quotient rule
(u/v)′ = (u′v − uv′)/v²
Chain rule
f(g(x))′ = f′(g(x))·g′(x)
Common derivatives
(eˣ)′ = eˣ, (ln x)′ = 1/x, (sin x)′ = cos x, (cos x)′ = −sin x

Method

  1. Identify the structure: sum, product, quotient or composition.
  2. Apply the power rule to each power of x.
  3. Use the product, quotient or chain rule where needed.
  4. Simplify, and evaluate at the point if asked.

Integrals

Power rule
∫xⁿ dx = xⁿ⁺¹/(n + 1) + C
Definite integral
∫ab f dx = F(b) − F(a)
Common integrals
∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
Area between curves
∫ab (f − g) dx

Method

  1. Rewrite the integrand as a sum of simple terms where possible.
  2. Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for ex, 1/x, sin and cos.
  3. Use substitution when you see a function and its derivative together.
  4. For a definite integral, evaluate F(b) − F(a).

Complex numbers

Rectangular form
z = a + jb, j² = −1, z̄ = a − jb
Modulus and argument
|z| = √(a² + b²), tan θ = b/a (check the quadrant)
Polar and exponential form
z = r∠θ = r(cos θ + j sin θ) = r·ejθ
Division
z₁/z₂ = z₁z̄₂/|z₂|²
Product and quotient
r₁∠θ₁ · r₂∠θ₂ = r₁r₂∠(θ₁ + θ₂), r₁∠θ₁ / r₂∠θ₂ = (r₁/r₂)∠(θ₁ − θ₂)
De Moivre
(r∠θ)ⁿ = rⁿ∠nθ
n-th roots
zk = R1/n∠((φ + 360°·k)/n), k = 0 … n − 1
Quadratic with p² < 4q
z = −p/2 ± j√(q − p²/4)

Method

  1. Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
  2. Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
  3. Convert with r = √(a² + b²) and θ from a sketch of the point (check the quadrant), and back with a = r cos θ, b = r sin θ.
  4. For powers use De Moivre, (r∠θ)ⁿ = rⁿ∠nθ; for zⁿ = R∠φ list all n roots R1/n∠((φ + 360°k)/n).

Differential equations

Exponential growth/decay
dy/dt = ky → y = y0ekt
Half-life
k = ln 2 / t½
Newton's law of cooling
T(t) = Ta + (T0 − Ta)e−kt
Integrating factor
y′ + p(x)y = q(x): μ = e∫p dx, (μy)′ = μq
Mixing tank
dm/dt = r·c − (r/V)·m

Matrices & determinants

2×2 determinant
det = ad − bc
2×2 inverse
A⁻¹ = (1/det)·[d −b; −c a]
Product
(AB)ij = Σk AikBkj
Eigenvalues
det(A − λI) = 0
Cramer’s rule
xi = det(Ai)/det(A)

Vectors: dot & cross product

Magnitude
|a| = √(ax² + ay² + az²)
Unit vector
û = a/|a|
Components from an angle
Fx = F·cos θ, Fy = F·sin θ
Dot product
a·b = axbx + ayby + azbz = |a||b|cos θ
Scalar projection of a onto b
a·b/|b|
Work
W = F·d
Cross product
a × b = (aybz − azby, azbx − axbz, axby − aybx)
Areas
parallelogram |a × b|, triangle ½|a × b|
Moment about O
MO = r × F, r = A − O
Volume of a parallelepiped
V = |a·(b × c)|

Method

  1. Write every vector in components; for a force given by a magnitude and an angle, use F·cos θ and F·sin θ.
  2. For lengths and directions use the magnitude √(x² + y² + z²) and the unit vector a/|a|.
  3. Use the dot product for angles, projections, perpendicularity and work: a·b = |a||b|cos θ.
  4. Use the cross product for perpendicular vectors, areas and moments: write the three components carefully, keeping the order a × b.

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

2Practice problems

1.LimitsMedium
xyy = L

Find the limit L = limx→∞ (6x3 + 4)/(3x3 + 2x + 8).

f(x) = (6x3 + 4)/(3x3 + 2x + 8)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (e−0.5x − 1 + 0.5x)/x².

f(x) = (e−0.5x − 1 + 0.5x)/x²x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x·sin(3x), with x in radians. Find f′(1.5).

f(x) = x·sin(3x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
xy

Let f(x) = x3 − 6x2 + 9x − 2. Find the x-coordinate x of the local maximum.

f(x) = x3 − 6x2 + 9x − 2
x =
5.IntegralsMedium
xy14y = x ln x

Evaluate I = ∫₁4 x·ln x dx.

∫₁4 x ln x dx
I =
6.IntegralsHard
xyy = x2​y = 4x

Find the area A of the region enclosed by y = x2 and y = 4x.

y = x2y = 4x
A =
7.Complex numbersMedium
ReIm11−1−122−2−233−3−3θz

The number z = 3·ej1.2 has its angle in radians. Find its real part a.

r = 3θ = 1.2 rad
a =
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

One root of z³ − 11z² + 41z − 51 = 0 is z₁ = 4 + j. The coefficients are real. Find the real root r.

z₁ = 4 + j
r =
9.Differential equationsMedium
t (min)T (°C)17Ta​177 °C

A forged steel part at 380 °C is left to cool in a workshop at 26 °C. After 17 min its temperature is 177 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 380 °CTa = 26 °CT(17) = 177 °C
k = min⁻¹
10.Differential equationsHard
ty2y(2) = ?

Solve y′ + 2y = 5e−0.5t with y(0) = 2. Find y(2).

y′ + 2y = 5e−0.5ty(0) = 2
y(2) =
11.Matrices & determinantsMedium
A =−134−3−42−4−43

Find the determinant det A of the 3×3 matrix A shown.

A = [−1 3 4; −3 −4 2; −4 −4 3]
det A =
12.Matrices & determinantsHard
A =−1−6−21b =−1112

Solve the system −x − 6y = −11 and −2x + y = 12. Find x.

−x − 6y = −11−2x + y = 12
x =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (−4, −3, −3) and b = (1, −2, −1), in degrees.

a = (−4, −3, −3)b = (1, −2, −1)
θ = °
14.Vectors: dot & cross productHard
xyzab

Find the area A of the parallelogram spanned by a = (0, 2, 6) and b = (0, −1, 4).

a = (0, 2, 6)b = (0, −1, 4)
A =
15.ProbabilityMedium
xp(x)100.35200.25300.15400.25μ = 23

For the distribution shown, find the standard deviation σ of X.

x: 10, 20, 30, 40p(x): 0.35, 0.25, 0.15, 0.25
σ =
16.ProbabilityHard
n = 7, p = 0.25: P(X = 2) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 2 correct answers?

n = 7p = 0.25k = 2
P =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 59, 11, 23, 18, 22, 27.

n = 6
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is above 20?

μ = 60σ = 20x = 20
P =
19.Normal distributionMedium
48 kg47.2 kgT = X₁ + … + X₄?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 4 of them exceeds 47.2 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 4> 47.2 kg
P =
20.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.05 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =

3Solutions

1.Limits

Answer: L = 2

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

Divide by the highest power

(6 + 4/x3)/(3 + 2/x2 + 8/x3)

Let x → ∞

L = 6/3 = 2

Answer

L = 2

For equal degrees the limit is the ratio of the leading coefficients.

2.Limits

Answer: L = 0.125

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

First application

(−0.5·e−0.5x + 0.5)/(2x) → 0/0

Second application

0.25·e−0.5x/2

Substitute x = 0

L = 0.25·1/2 = 0.125

Answer

L = 0.125
3.Derivatives

Answer: f′(1.5) = −1.926

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Product and chain rule

f′(x) = sin(3x) + 3x·cos(3x)

Substitute

f′(1.5) = −0.9775 + 4.5·(−0.2108) = −1.926

Answer

f′(1.5) = −1.926
4.Derivatives

Answer: x = 1

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Derivative

f′(x) = 3x2 − 12x + 9 = 3·(x2 − 4x + 3)

Solve f′(x) = 0

x = 1 or x = 3

Which one is which

f″(x) = 6x − 12: negative at a maximum, positive at a minimum.

f″(1) = −6, f″(3) = 6
x = 1

Answer

x = 1
5.Integrals

Answer: I = 7.34

Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.

Choose u and dv

u = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(8·ln 4 − 4) − (0 − 0.25) = 7.34

Answer

I = 7.34
6.Integrals

Answer: A = 10.67

Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.

Intersection points

x2 = 4x → x = 0, x = 4

Integrate upper minus lower

A = [2x2 − 0.3333x3]04 = 10.67

Answer

A = 10.67
7.Complex numbers

Answer: a = 1.087

Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.

Euler's formula

z = 3·(cos 1.2 + j sin 1.2)

Components (radians)

a = 3·cos 1.2 = 1.087
b = 3·sin 1.2 = 2.796
a = 1.087

Answer

a = 1.087
8.Complex numbers

Answer: r = 3

De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.

Conjugate pair

z₂ = 4 − j
(z − z₁)(z − z₂) = z² − 2·4·z + (4² + 1²) = z² − 8z + 17

Factor out the quadratic

z³ − 11z² + 41z − 51 = (z² − 8z + 17)(z − 3)

Real root

r = 3

Answer

r = 3

Check with the sum of the roots: z₁ + z₂ + r = 8 + 3 = 11 = −(−11).

9.Differential equations

Answer: k = 0.05012 min⁻¹

First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.

Ratio of temperature differences

e−17k = 151/354

Solve for k

k = ln(354/151)/17 = 0.05012 min⁻¹

Answer

k = 0.05012 min⁻¹
10.Differential equations

Answer: y(2) = 1.202

First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.

Integrating factor

μ = e2t
(e2ty)′ = 5e1.5t

General solution

y = 5/1.5·e−0.5t + C·e−2t = 3.333·e−0.5t + C·e−2t

Initial condition

2 = 3.333 + C → C = −1.333

Evaluate

y(2) = 3.333·0.3679 + (−1.333)·0.01832 = 1.202

Answer

y(2) = 1.202
11.Matrices & determinants

Answer: det A = −9

Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.

2×2 minors of the first row

M₁₁ = −4, M₁₂ = −1, M₁₃ = −4

Cofactor expansion

det A = (−1)·(−4) − 3·(−1) + 4·(−4) = −9

Answer

det A = −9
12.Matrices & determinants

Answer: x = −4.692

Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.

Determinant of the coefficient matrix

det A = (−1)·1 − (−6)·(−2) = −13

Replace the x column

det Aₓ = (−11)·1 − (−6)·12 = 61
x = 61/(−13) = −4.692

Answer

x = −4.692
y = 2.615
13.Vectors: dot & cross product

Answer: θ = 69.51 °

Dot product: a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.

Dot product

a·b = (−4)·1 + (−3)·(−2) + (−3)·(−1) = 5

Magnitudes

|a| = √((−4)² + (−3)² + (−3)²) = 5.831
|b| = √(1² + (−2)² + (−1)²) = 2.449

Angle

cos θ = 5/(5.831·2.449) = 0.3501
θ = arccos(0.3501) = 69.51°

Answer

θ = 69.51°
14.Vectors: dot & cross product

Answer: A = 14

Cross product: a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.

Cross product

(a × b)x = aybz − azby = 2·4 − 6·(−1) = 14
(a × b)y = azbx − axbz = 6·0 − 0·4 = 0
(a × b)z = axby − aybx = 0·(−1) − 2·0 = 0

Area

A = |a × b| = √(14² + 0² + 0²) = 14

Answer

A = 14
15.Probability

Answer: σ = 11.87

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 10·0.35 + 20·0.25 + 30·0.15 + 40·0.25 = 23

Mean of the squares

E(X²) = 100·0.35 + 400·0.25 + 900·0.15 + 1600·0.25 = 670

Variance

Var(X) = 670 − 23² = 141
σ = √141 = 11.87

Answer

σ = 11.87
16.Probability

Answer: P = 0.3115

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Binomial formula

P(X = 2) = 21·0.252·0.755 = 21·0.0625·0.2373 = 0.3115

Answer

P = 0.3115
17.Descriptive statistics

Answer: s = 16.74

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 160/6 = 26.67

Squared deviations

Σ(x − x̄)² = 32.33² + (−15.67)² + (−3.667)² + (−8.667)² + (−4.667)² + 0.3333² = 1401

Standard deviation

s = √(1401/5) = 16.74

Answer

s = 16.74
18.Descriptive statistics

Answer: P = 0.9772

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = −2

Standard normal table

Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
19.Normal distribution

Answer: P = 0.6915

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the total

μT = 48 kg, σT = 0.8·√4 = 1.6 kg

Standardise

z = (47.2 − 48)/1.6 = −0.5

Probability

P = 1 − Φ(−0.5) = 0.6915

Answer

P = 0.6915
20.Normal distribution

Answer: P = 0.9909

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the difference

μ = 25.05 − 25 = 0.05 mm, σ = 0.02121 mm

Standardise

z = (0 − 0.05)/0.02121 = −2.36

Probability

P = 1 − Φ(−2.36) = 0.9909

Answer

P = 0.9909

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